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Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

I can't understand the solution it gave

Could someone explain how this getSum function works?

Here is the answer in JS:

var getSum=function(a,b) {
    const Sum = a^b; //I can't understand how those two line's code can
    const carry = (a & b) << 1; //get the sum
        if(!carry) {
            return Sum
    return getSum(Sum,carry);
console.log(getSum(5,1));

It's basically replicating the half-adder

Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below

╔═══════╤═════════════╗
║ Input │   Output    ║
╠═══╤═══╪═══════╤═════╣
║ A │ B │ carry │ sum ║
╟───┼───┼───────┼─────╢
║ 0 │ 0 │   0   │  0  ║
╟───┼───┼───────┼─────╢
║ 1 │ 0 │   0   │  1  ║
╟───┼───┼───────┼─────╢
║ 0 │ 1 │   0   │  1  ║
╟───┼───┼───────┼─────╢
║ 1 │ 1 │   1   │  0  ║
╚═══╧═══╧═══════╧═════╝
From the table we get the logic for the outputs: carry = A and B, sum = A xor B
XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside
So the snippet above is working like this
const Sum=a^b;              // sum = a xor b = a ⊕ b
const carry=(a&b)<<1;       // carry = 2*(a and b), since we carry to the next bit
if(!carry){
    return Sum;             // no carry, so sum + carry = sum
return getSum(Sum,carry);   // a + b = sum + carry
So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry
See also
Adding two numbers without + operator (Clarification)
What is the best way to add two numbers without using the + operator?
adds two numbers without using + or any arithmetic operators
Adding two numbers without using the addition operator
                If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.
– Sergiy Kolodyazhnyy
                Mar 16, 2019 at 12:04
                @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut
– phuclv
                Mar 16, 2019 at 12:14
                Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.
– ilkkachu
                Mar 16, 2019 at 21:59
Let's learn by example. Imagine that a = 3 and b = 5
In binary notation they are a = 0011 and b = 0101
a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11

So when we're doing a^b result will be 0110.
AND + SHIFT
a&b performs logical AND operation. It returns 1 only when a = b = 1.
In our case the result is 0001 
<< shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).
Next iterations:
Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)
Now a^b = 0100 and (a&b)<<1 = 0100
Repeating again.
Now a^b = 0000 and (a&b)<<1 = 1000 
And again.
Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8. 
Everything worked fine since 3+5=8
                This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?
– ilkkachu
                Mar 16, 2019 at 22:01
                @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.
– flppv
                Mar 17, 2019 at 0:35
 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
int carry = (p & q) << 1; // Left Shift, 1+1=2
if (carry != 0) {
    return getSum(result, carry);
return result;
Start By p=5,q=6. Then the XOR would be,
------
So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,
-------
We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get
 0100<<1=1000
So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.
getSum(3, 8);
So, doing the first XORing we get,
-------
The XORing this time yielded in 11 (1011 binary),so we perform the AND now,
-------
We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).
Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations. 
^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.
Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.
So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) { return getSum(a ^ b, (a & b) << 1); }, except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.
        
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